Let $R$ be the region enclosed by $y=\dfrac{x^3}{2}$, the line $y=4$, and the $y$ -axis. $y$ $x$ $ 4$ $ 2$ $ R$ ${y=\dfrac{x^3}{2}}$ Region $R$ is the base of a solid. For each $y$ -value, the cross section of the solid taken perpendicular to the $y$ -axis is a rectangle whose base lies in $R$ and whose height is $\dfrac y2$. Which one of the definite integrals gives the volume of the solid? Choose 1 answer: Choose 1 answer: (Choice A) A $\int_0^4 \sqrt[3]{2y}\cdot \dfrac y2\,dy$ (Choice B) B $\int_0^4 \left(4-\dfrac{y^3}{2}\right)\cdot \dfrac y2\,dy$ (Choice C) C $\int_0^2 \sqrt[3]{2y}\cdot \dfrac y2\,dy$ (Choice D) D $\int_0^2 \left(4-\dfrac{y^3}{2}\right)\cdot \dfrac y2\,dy$
Let's imagine the solid is made out of many thin slices that are perpendicular to the $y$ -axis. $y$ $x$ $ 4$ $ 2$ Each slice is a prism. Let the width of each slice be $dy$ and let the area of the prism's face, as a function of $y$, be $A(y)$. Then, the volume of each slice is $A(y)\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b A(y)\,dy$ What we now need is to figure out the expression of $A(y)$ and the interval of integration. Let's consider one such slice. $y$ $x$ $ 4$ $ 2$ ${y=\dfrac{x^3}{2}}$ $ b(y)$ $ h(y)$ $ dy$ $ A(y)$ The face of that slice is a rectangle with base $b(y)$ and height $h(y)$. We are given that $h(y)=\dfrac y2$. The base $b(y)$ is equal to the distance from the curve $y=\dfrac{x^3}{2}$ to the $y$ -axis. To find an expression for $b(y)$, we must rewrite $y=\dfrac{x^3}{2}$ as a function of $y$ (which means we need to isolate $x$ ). $x=\sqrt[3]{2y}$ So, for each $y$ -value, this is the base of the rectangle: $b(y)=\sqrt[3]{2y}$ Now we can find the area of the rectangle: $\begin{aligned} &\phantom{=}A(y) \\\\ &=b(y)h(y) \\\\ &=\sqrt[3]{2y}\cdot \dfrac y2 \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=4$. So the interval of integration is $[0,4]$. Now we can express the definite integral in its entirety! $\int_0^4 \sqrt[3]{2y}\cdot \dfrac y2\,dy$